What is the number of families of $4t+1$ subsets of order $2t$ of the set $\{1,2, \ldots ,p\}$, where $p$ is a prime number which is equal to $4t+1$ and the order of intersection of each pair of the subsets is $t$ or $t-1$ and each subset has $t$-intersection with $2t$ subsets and $t-1$-intersection with the other $2t$ subsets?

2 Answers
2

Yes, equivalently, the problem can be stated as follows: We are looking for the number of $p\times p$ 0-1-matrices $A$ with all row sums $2t$ such that $AA^T=tJ+tI-B$, where $B$ a 0-1-matrix with also all row sums equal to $2t$. (Here, $J$ denotes the all-1-matrix).

It is not hard to find circulant matrices that fulfill the conditions for each $p=4t+1$ (why only for primes?), and each of them gives rise to $(p-1)!/2$ set families. But different matrices can yield the same families, e.g. for $t=1$, we can have
$A=circ(11000)$ with $AA^T=circ(21001)$ and
$A=circ(10100)$ with $AA^T=circ(20110)$, each yielding the same 12 structures.

In fact, looking at the $t=2$ patterns that seem "complete", I'd conjecture that there are $\binom{2t}{t}$ essentially different such circulant matrices.
(There may be also non-circulant solutions, so this would not yet solve the problem, but be a nice start.)

A computer search for $t=3$ and $t=4$ shows that not all $\binom{2t}t$ patterns occur. E.g. for $t=3$, only 14 of the 20 occur, and on the other hand, some correspond to several non isomorphic matrices. E.g., both $A=circ(1111001000100)$ and $A=circ(1110110010000)$ yield $AA^T=circ(6\ 323322\ 223323)$.
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WolfgangFeb 24 '13 at 16:45

I do not know the answer. However, here is part of an
analysis for t=1; perhaps some of it can extend for
larger t. I usually mean set of size 2t when I say set.

Consider two sets with intersection C of size t. The complement
of their union has size t+1, which specifies only one set
which has intersection of size t-1 with the two sets.
This means any other set that intersects the two sets
misses C. Thus (as t=1) one can specify a structure by
listing the locations of successive C's. In this case, it
means enumerating 5 cycles and dividing out by
cyclic and reversal symmetry, giving 12 such structures.
(I view it as 0-1 matrices where each column is proved
to have 2 1's.)